Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{r + 4}{2r - 20} \times \dfrac{r^2 + r}{r^2 + 5r + 4} $
Solution: First factor the quadratic. $n = \dfrac{r + 4}{2r - 20} \times \dfrac{r^2 + r}{(r + 4)(r + 1)} $ Then factor out any other terms. $n = \dfrac{r + 4}{2(r - 10)} \times \dfrac{r(r + 1)}{(r + 4)(r + 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (r + 4) \times r(r + 1) } { 2(r - 10) \times (r + 4)(r + 1) } $ $n = \dfrac{ r(r + 4)(r + 1)}{ 2(r - 10)(r + 4)(r + 1)} $ Notice that $(r + 1)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ r\cancel{(r + 4)}(r + 1)}{ 2(r - 10)\cancel{(r + 4)}(r + 1)} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $n = \dfrac{ r\cancel{(r + 4)}\cancel{(r + 1)}}{ 2(r - 10)\cancel{(r + 4)}\cancel{(r + 1)}} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $n = \dfrac{r}{2(r - 10)} ; \space r \neq -4 ; \space r \neq -1 $